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Basile [38]
2 years ago
11

A box is at rest on a slope

Physics
1 answer:
matrenka [14]2 years ago
6 0

Answer:

170 N

Explanation:

The following data were obtained from the question:

Mass (m) = 20 Kg

Angle (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Weight (W) =?

The perpendicular component of the weight can be obtained by using the following formula:

W = mg × Cosθ

W = 20 × 9.8 × Cos 30

W = 196 × 0.866

W = 169.7 ≈ 170 N

Therefore, perpendicular component of the weight is 170 N

You might be interested in
A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

\mu = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N

Magnitude of frictional force is 628.022466 N

F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2

v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s

Initial speed of the player is 8.61 m/s

4 0
3 years ago
How much work is done if a force of 20 N moves an object a distance of 6 m?​
Nesterboy [21]

Explanation:

W=F×s

w=20N×6m=120J

W=120J

3 0
2 years ago
When u write on a piece of glass sheet with a piece of chalk , the writing is not clear explain .
svetlana [45]
Becuse your weighting with chalk that has pigment
6 0
3 years ago
Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very
timofeeve [1]

Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ v=\frac{502.1}{\sqrt{3} }

      =289.9 \ m/s

The time will be:

⇒ t=\frac{d}{v}

      =\frac{2\times 6}{289.9}

      =\frac{12}{289.9}

      =0.041 \ sec

hence,

⇒ N=\frac{1}{t}

        =\frac{1}{0.041}

        =24.39 \ per \ sec

4 0
2 years ago
Find the area under the standard normal curve to the right of z=0.49. round your answer to four decimal places, if necessary.
VikaD [51]

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

<h3>What is  the standard normal curve?</h3>

The horizontal axis is approached by the standard normal bend as it extends indefinitely both in directions without ever being touched by it. The center of the bell-shaped, z=0 standard normal curve. Between z=3 and z=3, almost the entire area underneath the standard normal curve is located.

<h3>Use of the standard normal curve:</h3>

Use the normal distribution's standard form to calculate probability. Since the standard normal distribution is indeed a probability distribution, the probability that a variable will take on a range of values is indicated by area of the curve between two points. 100% or 1 is the total area beneath the curve.

<h3>According to the given data:</h3>

the region to the left of the standard normal curve,

z=0.49

To the right of,

z = 2.05

So,

The area will be:

= P[z < 0.49] + P[ z >2.05]

= P[z < 0.49] + 1 -  P[ z < 2.05]

= .6879 + 1 - .9798

= 0.7081

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

To know more about standard normal curve visit:

brainly.com/question/12972781

#SPJ4

I understand that the question you are looking for is:

Find the area under the standard normal curve to the left of z = 0.49 and to the right of z = 2.05. Round your answer to four decimal places, if necessary.

4 0
1 year ago
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